If you don't mind me asking: (a) why do you want to know? mere curiosity, or as a precursor to some other question? (b) what goes wrong with a direct attempt to use the definition of (sequential) compactness?
–
Yemon ChoiMar 14 '10 at 9:06

3 Answers
3

For a natural number $j$ let $f_j$ be the indicator function of the interval $[0,1/j]$ times the square root of $j$.
Then the $L^2$-norm of $f_j$ is one.
A simple calculation shows that one has
$||Tf_i-Tf_j||^2\ge\int_0^{1/j}(\sqrt i-\sqrt j)^2dx= (1-\sqrt{i/j})^2$ for $i\le j$, which implies that no subsequence can be Cauchy.

It is natural to look for an example around $x=0$ since that is where the kernel of $T$ fails to be $L^2$.

Anton already gave a very clean answer. Another way to see it is to work backwards: start from a sequence of functions $F_j$ in $L^2$ that does is non-compact, and define $f_j(x) = \frac{d}{dx} (x F_j(x) )$.

For example, let $\phi(x)$ be an arbitrary smooth bump function supported in $[-1/4,1/4]$, then the sequence of functions $F_j(x) = 2^j \phi( 4^j x - 1)$ all have disjoint support, but all have the same $L^2$ norm, so obviously does not have a converging subsequence in $L^2$.

Now set $f_j = (xF_j)' = F_j(x) + 8^j x \phi' (4^j x - 1)$. Since $\phi'$ has support only in $[-1/4,1/4]$, on the support of $f_j$ we can bound $4^j x$ absolutely by, say, 2. So we have that $f_j$ is a bounded sequence in $L^2$, whose corresponding $F_j = Tf_j$ cannot have a Cauchy subsequence.

Edit: I should also provide some motivation: observe that the scaling argument also works the other way (replace $j$ by $-j$, so that you can dilate). The Hardy-type inequality that you are using is a scaling invariant inequality: you estimate $f/x$ in $L^2$ by its derivative $f'$. If we treat $x$ as having units of distance, then the two objects have the same units regardless of what units $f$ has. This gives scaling invariance of the estimate. In other words, the estimate is invariant under the natural scaling action of $\mathbb{R}_{+}$ on $L^2(\mathbb{R}_+)$, where the group operation for $\mathbb{R}_{+}$ is multiplication.

Observe that $(\mathbb{R}_+, \times)$ is a non-compact Lie group. Generally, if you have an inequality/operator that is invariant under the action of a non-compact Lie group, the inequality/operator cannot be compact. You just need to start with some test function and act on it by the Lie group action to generate a bounded sequence that runs off non-compactly in the "infinity dimension" direction. Terry summarised it in his Buzz http://www.google.com/buzz/114134834346472219368/9UseDXTJN74/There-are-three-ways-that-sequential-compactness a short while back.

This is, of course, closely related to the notion of concentration compactness.

I took the liberty of editing your answer to solve the latex problem. This is a well-known problem and is caused by underscores and asterisks and you can solve it by enclosing the latex expression (including the dollar signs) in backticks.
–
José Figueroa-O'FarrillMar 14 '10 at 12:33

Let { $ L_{n} $ } be the sequence of Laguerre polynomials,
and let us define

$e_{n}(t)=\dfrac{L_{n}(\ln t)}{t}$ $\cdot\chi_{\left(1,\infty\right)}\left(t\right)$
$(n\in\mathbb{N\textrm{, t > 0}})$ . Then { $e_{n} $ } is
an orthonormal system in $L^{2}\left(0,\infty\right)$, and it is
not hard to see that $He_{n}=e_{n}-e_{n+1}$ $(n\in\mathbb{N})$,
where $H$ stands for the Hardy averaging operator. Therefore, $H$
cannot be compact.